When $K M n O_{4}$ acts as an oxidising agent and ultimately forms $[M n O_{4}]^{- 2}, M n O_{2}, M n_{2} O_{3}, M n^{+ 2}$ , then number of electrons transferred in each case is $p, q, r$ and $s$ respectively. Find the value of $(s - r)^{2} + (q - p)^{2}$ .

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Solution

Here the value of $p, q, r$ and $s$ respectively is $1, 3, 4$ and $5$ .
Hence $(s - r)^{2} + (q - p)^{2} = 1^{2} + 2^{2} = 5$

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When KMnO4 acts as an oxidising agent and ultimately forms [MnO4]−2, MnO2, Mn2O3, Mn+2, then number of electrons transferred in each case is p,q,r and s respectively. Find the value of (s−r)2+(q−p)2.

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When $K M n O_{4}$ acts as an oxidising agent and ultimately forms $[M n O_{4}]^{- 2}, M n O_{2}, M n_{2} O_{3}, M n^{+ 2}$ , then number of electrons transferred in each case is $p, q, r$ and $s$ respectively. Find the value of $(s - r)^{2} + (q - p)^{2}$ .

Here the value of $p, q, r$ and $s$ respectively is $1, 3, 4$ and $5$ .
Hence $(s - r)^{2} + (q - p)^{2} = 1^{2} + 2^{2} = 5$